The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(O(1), O(n^2)).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

din(der(plus(X, Y))) → u21(din(der(X)), X, Y)
u21(dout(DX), X, Y) → u22(din(der(Y)), X, Y, DX)
u22(dout(DY), X, Y, DX) → dout(plus(DX, DY))
din(der(times(X, Y))) → u31(din(der(X)), X, Y)
u31(dout(DX), X, Y) → u32(din(der(Y)), X, Y, DX)
u32(dout(DY), X, Y, DX) → dout(plus(times(X, DY), times(Y, DX)))
din(der(der(X))) → u41(din(der(X)), X)
u41(dout(DX), X) → u42(din(der(DX)), X, DX)
u42(dout(DDX), X, DX) → dout(DDX)

Rewrite Strategy: INNERMOST

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(U22(din(der(z2)), z1, z2, z0), DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(U32(din(der(z2)), z1, z2, z0), DIN(der(z2)))
U41(dout(z0), z1) → c7(U42(din(der(z0)), z1, z0), DIN(der(z0)))

S tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(U22(din(der(z2)), z1, z2, z0), DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(U32(din(der(z2)), z1, z2, z0), DIN(der(z2)))
U41(dout(z0), z1) → c7(U42(din(der(z0)), z1, z0), DIN(der(z0)))

K tuples:none
Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U31, U41

Compound Symbols:

c, c1, c2, c3, c5, c7

(3) CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID) transformation)

Removed 3 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

S tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

K tuples:none
Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U31, U41

Compound Symbols:

c, c1, c2, c3, c5, c7

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

We considered the (Usable) Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))

And the Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(DIN(x₁)) = [2]x₁   
POL(U21(x₁, x₂, x₃)) = 0   
POL(U31(x₁, x₂, x₃)) = [4]x₁   
POL(U41(x₁, x₂)) = [4]x₁   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(der(x₁)) = 0   
POL(din(x₁)) = 0   
POL(dout(x₁)) = [2] + x₁   
POL(plus(x₁, x₂)) = 0   
POL(times(x₁, x₂)) = [4]   
POL(u21(x₁, x₂, x₃)) = [2]x₁   
POL(u22(x₁, x₂, x₃, x₄)) = [4]x₁   
POL(u31(x₁, x₂, x₃)) = [4]x₁   
POL(u32(x₁, x₂, x₃, x₄)) = [2] + x₁ + x₄   
POL(u41(x₁, x₂)) = [4]x₁   
POL(u42(x₁, x₂, x₃)) = [3] + [3]x₁ + [2]x₃   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

S tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))

K tuples:

U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U31, U41

Compound Symbols:

c, c1, c2, c3, c5, c7

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

U21(dout(z0), z1, z2) → c3(DIN(der(z2)))

We considered the (Usable) Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))

And the Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(DIN(x₁)) = [1]   
POL(U21(x₁, x₂, x₃)) = [4]x₁   
POL(U31(x₁, x₂, x₃)) = [4]x₁   
POL(U41(x₁, x₂)) = x₁   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(der(x₁)) = 0   
POL(din(x₁)) = 0   
POL(dout(x₁)) = [2]   
POL(plus(x₁, x₂)) = [5]   
POL(times(x₁, x₂)) = [3]   
POL(u21(x₁, x₂, x₃)) = 0   
POL(u22(x₁, x₂, x₃, x₄)) = [4]x₁   
POL(u31(x₁, x₂, x₃)) = 0   
POL(u32(x₁, x₂, x₃, x₄)) = [4]x₁   
POL(u41(x₁, x₂)) = 0   
POL(u42(x₁, x₂, x₃)) = [4]x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

S tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))

K tuples:

U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))

Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U31, U41

Compound Symbols:

c, c1, c2, c3, c5, c7

(9) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))

We considered the (Usable) Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))

And the Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(DIN(x₁)) = x₁   
POL(U21(x₁, x₂, x₃)) = [2]x₁·x₃ + [2]x₁·x₂   
POL(U31(x₁, x₂, x₃)) = [2]x₁·x₃   
POL(U41(x₁, x₂)) = [2]x₁   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(der(x₁)) = x₁   
POL(din(x₁)) = 0   
POL(dout(x₁)) = [2] + x₁   
POL(plus(x₁, x₂)) = [2] + x₁   
POL(times(x₁, x₂)) = x₁   
POL(u21(x₁, x₂, x₃)) = [2]x₁ + [2]x₁·x₃ + [3]x₁²   
POL(u22(x₁, x₂, x₃, x₄)) = [2] + x₁ + x₄ + [3]x₄² + x₃·x₄   
POL(u31(x₁, x₂, x₃)) = [2]x₁·x₃ + [3]x₁·x₂   
POL(u32(x₁, x₂, x₃, x₄)) = [2]x₁ + [2]x₂ + [2]x₃·x₄ + [2]x₂·x₄   
POL(u41(x₁, x₂)) = [2]x₁·x₂   
POL(u42(x₁, x₂, x₃)) = [2]x₁ + [3]x₂ + [2]x₂·x₃   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

S tuples:

DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))

K tuples:

U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))

Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U31, U41

Compound Symbols:

c, c1, c2, c3, c5, c7

(11) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))

We considered the (Usable) Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))

And the Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(DIN(x₁)) = [1] + [2]x₁   
POL(U21(x₁, x₂, x₃)) = [1] + [2]x₁ + [2]x₁·x₃ + x₁²   
POL(U31(x₁, x₂, x₃)) = [1] + x₃ + [2]x₁·x₃   
POL(U41(x₁, x₂)) = [2]x₁²   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(der(x₁)) = x₁   
POL(din(x₁)) = 0   
POL(dout(x₁)) = [1] + x₁   
POL(plus(x₁, x₂)) = [1] + x₁   
POL(times(x₁, x₂)) = [1] + x₁ + x₂   
POL(u21(x₁, x₂, x₃)) = [2]x₁ + x₁·x₂   
POL(u22(x₁, x₂, x₃, x₄)) = [2] + [3]x₁ + x₂ + x₄ + [3]x₁²   
POL(u31(x₁, x₂, x₃)) = [2]x₁ + [3]x₁·x₃ + [3]x₁·x₂   
POL(u32(x₁, x₂, x₃, x₄)) = [1] + [2]x₂ + [2]x₃ + [2]x₁²   
POL(u41(x₁, x₂)) = [2]x₁·x₂   
POL(u42(x₁, x₂, x₃)) = x₂ + x₂·x₃ + [2]x₁²   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

S tuples:

DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))

K tuples:

U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))

Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U31, U41

Compound Symbols:

c, c1, c2, c3, c5, c7

(13) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^2))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))

We considered the (Usable) Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))

And the Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(DIN(x₁)) = [2] + x₁   
POL(U21(x₁, x₂, x₃)) = x₁·x₃ + x₁²   
POL(U31(x₁, x₂, x₃)) = x₃ + [2]x₁²   
POL(U41(x₁, x₂)) = [2]x₁²   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c2(x₁, x₂)) = x₁ + x₂   
POL(c3(x₁)) = x₁   
POL(c5(x₁)) = x₁   
POL(c7(x₁)) = x₁   
POL(der(x₁)) = [2] + x₁   
POL(din(x₁)) = 0   
POL(dout(x₁)) = [2] + x₁   
POL(plus(x₁, x₂)) = x₁   
POL(times(x₁, x₂)) = x₁ + x₂   
POL(u21(x₁, x₂, x₃)) = [2]x₁ + [2]x₁·x₃ + [3]x₁² + [2]x₁·x₂   
POL(u22(x₁, x₂, x₃, x₄)) = [3] + [3]x₁ + [3]x₂ + [3]x₃ + x₄ + x₃·x₄ + x₂·x₄   
POL(u31(x₁, x₂, x₃)) = [2]x₁·x₃ + [3]x₁² + [2]x₁·x₂   
POL(u32(x₁, x₂, x₃, x₄)) = [2]x₁ + [3]x₂ + [3]x₃ + [3]x₄ + x₃·x₄ + [2]x₂·x₄ + [3]x₁·x₄ + [3]x₁² + [2]x₁·x₂ + [3]x₁·x₃   
POL(u41(x₁, x₂)) = [2]x₁·x₂ + [2]x₁²   
POL(u42(x₁, x₂, x₃)) = [1] + [3]x₂ + [3]x₃ + x₃² + x₂·x₃ + x₁²   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

din(der(plus(z0, z1))) → u21(din(der(z0)), z0, z1)
din(der(times(z0, z1))) → u31(din(der(z0)), z0, z1)
din(der(der(z0))) → u41(din(der(z0)), z0)
u21(dout(z0), z1, z2) → u22(din(der(z2)), z1, z2, z0)
u22(dout(z0), z1, z2, z3) → dout(plus(z3, z0))
u31(dout(z0), z1, z2) → u32(din(der(z2)), z1, z2, z0)
u32(dout(z0), z1, z2, z3) → dout(plus(times(z1, z0), times(z2, z3)))
u41(dout(z0), z1) → u42(din(der(z0)), z1, z0)
u42(dout(z0), z1, z2) → dout(z0)

Tuples:

DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))

S tuples:none
K tuples:

U31(dout(z0), z1, z2) → c5(DIN(der(z2)))
U41(dout(z0), z1) → c7(DIN(der(z0)))
U21(dout(z0), z1, z2) → c3(DIN(der(z2)))
DIN(der(plus(z0, z1))) → c(U21(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(times(z0, z1))) → c1(U31(din(der(z0)), z0, z1), DIN(der(z0)))
DIN(der(der(z0))) → c2(U41(din(der(z0)), z0), DIN(der(z0)))

Defined Rule Symbols:

din, u21, u22, u31, u32, u41, u42

Defined Pair Symbols:

DIN, U21, U31, U41

Compound Symbols:

c, c1, c2, c3, c5, c7

(15) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(16) BOUNDS(O(1), O(1))